roots of equation.

[man111]

[tex]x^5+5\lambda x^4+(\lambda a-4)x^2-(8\lambda+3)x+\lambda a-2=0[/tex]. Then the value of [tex]a[/tex] for which the Given equation has one root Independent of [tex]\lambda[/tex].

[martin123456]

Just an idea, it's not finished.
Write the given equation as [tex]x^5-4x^2-3x-2=-\lambda(5x^4-ax^2+8x-a)[/tex].
Notice: I assume that [tex]a[/tex] is not dependent on [tex]\lambda[/tex]. But this might not be the case. Nevertheless,
let the equation has a root [tex]x_0[/tex], independent of [tex]\lambda[/tex], then the LHS is independent of [tex]\lambda[/tex] and then it's a must for the RHS to be independent too. So, having in mind that we assumed that [tex]a[/tex] is not dependent of [tex]\lambda[/tex] it's a must to have the following dependency: [tex]\exist x \in \mathbb{R}[/tex] that is a root of the LHS [tex]\Rightarrow[/tex] it's a root of the [tex]5x^4-ax^2+8x-a[/tex].